Identify the independent variable in this investigation (effect of wind speed on transpiration).

• The rate of water loss from the plant.
• The change in mass of the plant and pot over time.
• Wind speed.
• The type of plant used (Japanese spiraea).

Describe a method… to investigate the effect of wind speed on the rate of transpiration…

• Measure Transpiration: For each wind speed, record initial mass on a balance, expose plant to wind for a fixed time (e.g., 30 mins) using a stop-clock, record final mass, and calculate mass loss (water loss).
• Establish Wind Speeds: Use a variable-speed fan at different settings/distances, measuring each actual wind speed with an anemometer; include a ‘no wind’ (zero speed) control condition. Standardise the distance and angle of the fan relative to the plant.
• Replication: Repeat the measurement process (weigh, expose, reweigh) at least three times for each selected wind speed and calculate the mean rate of mass loss for each speed.
• Control Conditions: Place the potted plant (with soil surface sealed to prevent direct evaporation) in an environment where temperature and light intensity are kept constant and monitored throughout the experiment. Allow the plant to acclimatise to conditions before starting measurements.
• Watering: Ensure the soil is allowed to dry out completely between different wind speed trials to guarantee that water availability becomes the primary limiting factor, not wind.
• Safety: Keep hands and loose items clear of the fan blades; handle the plant and soil carefully (e.g., wear gloves if necessary, clean up spills).

Predict the effect of increasing wind speed on the rate of transpiration.

• Effect: Increasing wind speed generally increases the rate of transpiration, particularly at lower to moderate wind speeds.
• Explanation: Wind significantly increases the temperature of the leaf surface, causing water inside the leaf to evaporate more rapidly through the stomata.
• Explanation: Wind physically blows away the layer of humid air (high water vapour concentration) that accumulates near the leaf surface, particularly around the stomata (the boundary layer).
• Explanation: By removing the humid boundary layer, wind maintains a steeper water potential gradient (or water vapour concentration gradient) between the moist air inside the leaf and the drier surrounding air, thus enhancing the rate of diffusion of water vapour out of the stomata.

For one leaf (leaf 2), the time taken for the cobalt chloride paper to change colour was 137 seconds. Calculate the rate of transpiration in units of h⁻¹ using the formula: Rate = 3600 / time taken in seconds. Give your answer to three significant figures.

• Calculation: Substitute the time into the formula: Rate = 3600 / 137.
• Result rounded to 3 significant figures: 26.3 h⁻¹
• Result unrounded (intermediate step): 26.27737… h⁻¹
• Calculation: Incorrectly invert the formula: Rate = 137 / 3600.

Besides testing more leaves, state one other change to this cobalt chloride paper method that could improve the validity or reliability of the results.

• Ensure the cobalt chloride paper makes consistent and full contact with the leaf surface each time it is applied (e.g., using clips carefully).
• Conduct the experiment in complete darkness to prevent photosynthesis from influencing stomatal aperture during the measurement.
• Use pieces of cobalt chloride paper that are cut to a standardised size/area for every measurement.
• Use a standardised colour chart or a colorimeter (if feasible) to judge the colour change endpoint more objectively and consistently.
• Select leaves of a similar age, size, and position on the plant for comparison to minimize biological variation.
• Keep environmental conditions (light intensity, temperature, ambient humidity) as constant as possible during the measurement period.

State three environmental conditions in the glasshouse that should be standardised during the initial 10-week acclimatisation period, other than temperature.

• Humidity / Relative humidity / Air moisture content.
• The species of tree being used (e.g., isohydric vs anisohydric).
• Light intensity / Photosynthetically active radiation (PAR).
• Carbon dioxide concentration in the air.
• Duration of the light period per day / Photoperiod.
• Volume and/or frequency of watering / Soil moisture level.
• Soil type / Potting medium / Soil pH / Volume of soil.

Data compared well-watered trees at 28°C (Group 1) vs 35°C (Group 2). Isohydric species: mean conductance ~470 (G1) vs ~400 (G2) mmol m⁻² s⁻¹. Anisohydric species: ~280 (G1) vs ~250 (G2). State the effect of the high mean temperature (35°C) on the mean stomatal conductance of these well-watered young trees.

• The high temperature (35°C) caused a large increase in the mean stomatal conductance, especially in the isohydric species.
• Overall, the high temperature (35°C) resulted in a lower mean stomatal conductance compared to the lower temperature (28°C).
• This effect of reduced stomatal conductance at the higher temperature was observed in both the isohydric and the anisohydric species.

Data compared trees under water stress at 28°C (G3) vs 35°C (G4). Iso: ~20 (G3) vs ~70 (G4). Aniso: ~75 (G3) vs ~70 (G4). Difference G3 vs G4 significant for Iso (p=0.046), not for Aniso (p=0.788). Suggest and explain conclusions.

• Isohydric Response: Stomatal conductance significantly increased when high temperature was combined with water stress (mean increased from ~20 at 28°C to ~70 at 35°C).
• Anisohydric Response: Stomatal conductance remained low (~70-75) and did not show a statistically significant change when high temperature was imposed on water-stressed plants.
• Explanation (Isohydric): The increase suggests isohydric species might prioritise preventing potentially lethal high leaf temperatures (via evaporative cooling from transpiration) over strict water conservation when faced with combined heat and moderate water stress.
• General Water Stress Effect: The data confirms that water stress generally causes stomata to open wider compared to well-watered conditions to maximize CO₂ intake needed for survival.
• Explanation (Anisohydric): Their lack of significant response might indicate they prioritise maintaining some level of CO₂ uptake under stress, or that their stomata were already near minimum aperture due to the initial water stress, leaving little room for further response to temperature.

State a suitable null hypothesis for the t-test comparing the percentage of dead leaves in isohydric vs anisohydric species under water stress and high temperature.

• There is a significant difference between the mean percentage of dead leaves for isohydric species and the mean percentage for anisohydric species under these stress conditions.
• Isohydric species will have a significantly higher mean percentage of dead leaves than anisohydric species when exposed to water stress and high temperature.
• There is no significant difference between the mean percentage of leaves that were dead on young trees from isohydric species and young trees from anisohydric species when exposed to water stress and high temperature.

The denominator value for the t-test √(s₁²/n₁ + s₂²/n₂) was calculated as 8.231. Using the mean values provided (Isohydric mean = 19.0%, Anisohydric mean = 3.5%), calculate the value of t.

• Calculation step: The numerator is the difference between the means: |19.0 – 3.5| = 15.5.
• Formula uses sum of means: The t-value is calculated as (19.0 + 3.5) / 8.231.
• Result: The calculated t-value is t = 15.5 / 8.231 ≈ 1.88.
• Formula uses difference of means: The general formula for an independent samples t-test statistic is t = (Difference between sample means) / (Standard error of the difference between means), which corresponds to |x̄₁ – x̄₂| / denominator.

The calculated t-value is 1.883. The critical value for p=0.05 with the relevant degrees of freedom (48) is 2.011. Describe the conclusion from this t-test.

• The calculated t-value (1.883) is greater than the critical value (2.011) at p=0.05.
• Therefore, the difference between the means is statistically significant at the p=0.05 level.
• The null hypothesis (of no significant difference) is accepted (or rather, we fail to reject it).
• The comparison shows that the calculated t-value (1.883) is less than the critical value (2.011).
• Conclusion: There is no statistically significant difference (at the 5% level) in the mean percentage of dead leaves between the isohydric and anisohydric species under these conditions.

State a suitable control for the mosquito repellent investigation (testing attraction to a treated arm).

• Testing the same set of repellents on a different species of mosquito, such as Anopheles, which transmits malaria.
• Using a mechanical ‘artificial arm’ device coated with a standardised synthetic human scent instead of a live human volunteer’s arm.
• Having the same human volunteer perform the experiment identically but with no repellent applied to their arm (or perhaps applying only the solvent/carrier used for the repellents).
• Increasing the number of mosquitoes released into the cage for each trial from 200 to 500 to get clearer results.

Suggest one potential risk to the scientists or volunteer when carrying out this mosquito investigation and state a suitable precaution.

• Risk: Mosquito bites occurring despite the repellent or during handling. Precaution: Ensure the cage is securely sealed, volunteer wears protective clothing on unexposed areas, use aspirators carefully.
• Risk: Significant eye strain from prolonged observation and counting of mosquitoes in the cage sections. Precaution: Use magnification and automated counting software if possible, take frequent breaks, ensure good ergonomic setup and lighting.
• Risk: Allergic reaction or skin irritation caused by one of the repellent chemicals applied to the arm. Precaution: Ensure good laboratory ventilation, perform a patch test on the volunteer beforehand, provide access to washing facilities, consider use of gloves/mask when handling concentrated repellents.
• Risk: Accidental transmission of mosquito-borne diseases (e.g., dengue, Zika via Aedes aegypti) if lab strains are contaminated or safety protocols fail. Precaution: Use certified pathogen-free lab-reared mosquitoes, maintain strict containment procedures, wear appropriate protective clothing.

Results (% mosquitoes attracted, mean±SE): 40%DEET=68.6±6.4; 98%DEET=33.7±4.1; 30%LemonEucalyptus=29.6±6.3; 10%Picaridin=78.7±6.0. Student concluded DEET/LE were best & chose DEET for malaria. Evaluate this.

• Limitation: The experiment measured the percentage of mosquitoes attracted to a section near the arm, which may not directly correlate with the number of actual bites that transmit disease.
• Support: The data shows that 98% DEET and 30% Lemon Eucalyptus resulted in the lowest mean percentages of attracted mosquitoes, suggesting they were the most effective repellents *in this specific test*.
• Limitation: The mosquito species tested was Aedes aegypti (a vector for dengue, Zika, etc.), whereas malaria is transmitted by Anopheles species. Repellent effectiveness can vary between species, so these results are not directly applicable to malaria prevention.
• Support: The student’s decision is strongly supported because DEET is widely recognized by health authorities as an effective repellent against Anopheles mosquitoes that transmit malaria.
• Limitation: The standard error ranges for 98% DEET (33.7±4.1 spans ~29.6 to 37.8) and 30% Lemon Eucalyptus (29.6±6.3 spans ~23.3 to 35.9) overlap considerably, suggesting the observed difference in their mean effectiveness might not be statistically significant based on this data alone.
• Limitation: The experiment likely used only one volunteer (not specified but common in such setups), meaning the results might reflect that individual’s specific attraction level and skin chemistry, and may not be generalisable to other people.
• Limitation: The data shows 40% DEET was relatively ineffective (high attraction). Choosing “DEET” without specifying the concentration (98% was effective, 40% was not) makes the student’s conclusion imprecise based on the results.